If the solution curve f(x, y)=0 of the differential equation [(1+ _e x) (d x)/(d y)-x _e x…

Question

If the solution curve $f(x, y)=0$ of the differential equation

$$\left(1+\log _{e} x\right) \frac{d x}{d y}-x \log _{e} x=e^{y}, x > 0$$,

passes through the points $(1,0)$ and $$(\alpha, 2)$$, then $$\alpha^{\alpha}$$ is equal to :

Accepted Answer

We have, $\left(1+\log _e x\right) \frac{d x}{d y}-x \log _e x=e^{y}, x>0$

Put $x \log _e x=t$

$$
\begin{aligned}
& \Rightarrow  \left(x \cdot \frac{1}{x}+\log _e x\right) \frac{d x}{d y}  =\frac{d t}{d y} \\
& \Rightarrow  \left(1+\log _e x\right) \frac{d x}{d y}  =\frac{d t}{d y} \\
& \therefore  \frac{d t}{d y}-t  =e^{y}
\end{aligned}
$$

Now, IF $=e^{\int-d y}=e^{-y}$

$\begin{aligned} & \therefore \text { General solution, } t\left(e^{-y}\right)=\int\left(e^y \cdot e^{-y}\right) d y+c \\ & \Rightarrow t e^{-y}=\int d y+c \\ & \Rightarrow t e^{-y}=y+c \\ & \Rightarrow \left(x \log _e x\right) e^{-y}=y+c ..........(i)\end{aligned}$

Equation (i), passes through the point $(1,0)$

$$
\begin{aligned}
& \therefore 0=0+c \\
& \Rightarrow c=0 \\
& \therefore \left(x \log _e x\right) e^{-y}=y \\
& \Rightarrow x \log _e x=y e^y
\end{aligned}
$$

Which passes through $(\alpha, 2)$

$$
\begin{aligned}
& \therefore \alpha \log _e \alpha=2 e^2 \\
& \Rightarrow \log _e \alpha^\alpha=2 e^2 \\
& \Rightarrow \alpha^\alpha=e^{2 e^2}
\end{aligned}
$$

Solutions · 1